JAMB UTME - Chemistry - 2014
Tambaya 1 Rahoto
H 2O (g) + C (s) → H2(g) + CO(g)The oxidizing agent in the reaction above is
Tambaya 2 Rahoto
The process of converting linear alkanes to branched chain and cyclic hydrocarbons by heating in the presence of a catalyst to improve the quality of petrol is referred to as
Bayanin Amsa
The process described is called "reforming." Reforming is a process used in the petroleum industry to convert straight-chain hydrocarbons, which have low octane ratings and do not burn efficiently, into branched chain or cyclic hydrocarbons, which have higher octane ratings and burn more efficiently. The process involves heating the hydrocarbons in the presence of a catalyst, which causes the molecules to rearrange into a more desirable form. The resulting product is then used to improve the quality of gasoline, making it burn more efficiently and produce less harmful emissions.
Tambaya 3 Rahoto
In an equilibrium reaction, which of the following conditions indicates that maximum yield of the product will be obtained?
Tambaya 4 Rahoto
CH3COOC2H5(1) + H2O(1) ⇌ CH 3COOH (aq) + C 2H 5OH (aq)The purpose of H+ in the reaction above is to
Bayanin Amsa
Tambaya 5 Rahoto
Cr2O2−7 (aq) + H2O(1)⇌ 2CrO 2−4 (aq) + 2H + (aq)
What happens to the reaction above when the hydrogen ion concentration is increased?
Bayanin Amsa
In the given chemical reaction, Cr2O2−7 ions react with water to form 2CrO 2−4 ions and hydrogen ions (H+). This reaction is in equilibrium, meaning the reactants are forming products and the products are forming reactants at the same rate. According to Le Chatelier's principle, when a stress is applied to a system in equilibrium, the system will shift to counteract the stress. In this case, increasing the hydrogen ion concentration will increase the stress on the equilibrium system. To counteract this stress, the system will shift to consume the additional hydrogen ions. Since hydrogen ions are one of the products of the reaction, the system will shift to form more reactants and less products. Therefore, the equilibrium position will shift to the left. So, the correct answer is (D) the equilibrium position will shift to the left.
Tambaya 6 Rahoto
The gas that is removed by the water in the flask is
Tambaya 9 Rahoto
On exposure to the atmosphere, a hydrated salt loses its water of crystallization to become anhydrous. This phenomenon is referred to as
Bayanin Amsa
The phenomenon in which a hydrated salt loses its water of crystallization when exposed to the atmosphere is called efflorescence. Hydrated salts are those that have water molecules chemically bound to their crystal structures. When exposed to the atmosphere, the water molecules in the hydrated salt can evaporate into the air, leaving behind a dry, anhydrous salt. This process occurs due to the difference in vapor pressure between the water molecules in the salt and the water molecules in the air. If the vapor pressure of the water molecules in the air is lower than that of the water molecules in the salt, the water molecules in the salt will evaporate into the air. Efflorescence is commonly observed in certain salts such as sodium carbonate (Na2CO3), which is a hydrated salt containing ten water molecules. When exposed to air, Na2CO3 loses its water molecules and becomes anhydrous. The anhydrous form of Na2CO3 is commonly known as soda ash. Therefore, the correct answer is option A) efflorescence.
Tambaya 10 Rahoto
A silicon -containing ore has 92% 28Si, 5% 29Si and 3% 30Si. Calculate the relative atomic mass of the silicon.
Bayanin Amsa
The relative atomic mass of silicon can be calculated as follows: Relative atomic mass of silicon = (92% * 28Si atomic mass) + (5% * 29Si atomic mass) + (3% * 30Si atomic mass) where the atomic masses of 28Si, 29Si, and 30Si are, respectively, 28, 29, and 30 atomic mass units (amu). So, the relative atomic mass of silicon = (92% * 28 amu) + (5% * 29 amu) + (3% * 30 amu) = (0.92 * 28 amu) + (0.05 * 29 amu) + (0.03 * 30 amu) = 25.76 + 1.45 + 0.9 = 28.11 amu Therefore, the relative atomic mass of silicon is 28.11 amu.
Tambaya 11 Rahoto
Which of the following will shift to the left from dilute tetraoxosulphate (VI) acid?
Bayanin Amsa
Dilute tetraoxosulphate (VI) acid is also known as sulfuric acid (H2SO4). When an acid dissociates, it produces hydrogen ions (H+) and the corresponding anions. The extent to which the acid dissociates is dependent on its strength. Strong acids dissociate completely in solution, while weak acids only partially dissociate. In the given options, magnesium is the only metal that will shift to the left from dilute tetraoxosulphate (VI) acid. This is because magnesium is more reactive than hydrogen, and will displace hydrogen from the acid to form magnesium sulfate and hydrogen gas. This displacement reaction will shift the equilibrium position to the left, reducing the concentration of hydrogen ions and increasing the concentration of undissociated acid. The other metals listed - lead, copper, and gold - are less reactive than hydrogen and will not displace hydrogen from the acid. Therefore, their equilibrium position will remain unchanged and they will not shift to the left or the right.
Tambaya 12 Rahoto
Due to the unstable nature of ethyne, it is stored by dissolving in
Tambaya 14 Rahoto
How many faraday of electricity is required to produce 0.25 mole of copper?
Bayanin Amsa
Cu → Cu2+ + 2e
1mole Cu requires 2f
0.25 mole Cu requires x
x = 0.25x2/1 = 0.50f
Tambaya 15 Rahoto
A hydrocarbon has an empirical formula CH and a vapour density of 39. Determine its molecular formula.[C = 12, H = 1]
Bayanin Amsa
molecular mass = 2 x vapour density
= 2 x 39 = 79
∴ Cn Hn = 78
12n + 1n = 78
13n = 78
n = 7813
= 6
∴ Cn Hn = C 6H 6
Tambaya 16 Rahoto
The nitrogen obtained from air has a density higher than the one from nitrogen-containing compounds because the one from air is contaminated with
Bayanin Amsa
Tambaya 17 Rahoto
Which of the following salts will turn blue litmus red?
Bayanin Amsa
Blue litmus paper is used to test whether a substance is acidic or not. If a substance is acidic, it will turn blue litmus paper red. Out of the given options, only one substance is acidic in nature, which is potassium hydrogentetraoxosulphate (lV). Sodium tetrahydroxozincate (ll), Sodium trioxocarbonate (lV), and Zinc chloride hydroxide are either neutral or basic in nature and will not turn blue litmus paper red. Therefore, the answer is potassium hydrogentetraoxosulphate (lV).
Tambaya 18 Rahoto
The petroleum fraction that is used in heating furnaces in industries is
Bayanin Amsa
Tambaya 19 Rahoto
If the change in free energy of a system is -899 Jmol-1 and the entropy change is 10Jmol-1k-1 at 250C, calculate the enthalpy change.
Bayanin Amsa
The enthalpy change can be calculated using the formula: ΔG = ΔH - TΔS where ΔG is the change in free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin. Plugging in the values given: -899 Jmol-1 = ΔH - (250+273)K * 10Jmol-1K-1 Solving for ΔH: ΔH = -899 Jmol-1 + (523K * 10Jmol-1K-1) ΔH = -899 Jmol-1 + 5230 Jmol-1 ΔH = +4331 Jmol-1 Therefore, the enthalpy change is +4331 Jmol-1, which corresponds to +2081 Jmol-1.
Tambaya 20 Rahoto
Zn (s) + CuSO 4(aq) → ZnSO 4(aq) + Cu (s)
In the reaction above, the oxidation number of the reducing agent changes from
Bayanin Amsa
In the given chemical reaction, Zn is the reducing agent because it causes the reduction of Cu²⁺ to Cu. Reduction is a process where an atom or ion gains electrons. In this reaction, Zn loses electrons and gets oxidized to Zn²⁺. The oxidation number of an atom or ion is the charge it would have if all its bonds were completely ionic. Before the reaction, the oxidation state of Zn is 0 because it is in its elemental form. After the reaction, Zn is oxidized to Zn²⁺, which has an oxidation state of +2. Therefore, the oxidation number of the reducing agent (Zn) changes from 0 to +2. So, the correct answer to the question is "0 to +2".
Tambaya 21 Rahoto
The mathematical expression of PH is
Tambaya 23 Rahoto
16.55g of lead (ll) trioxonitrate (V) was dissolved in 100g of distilled water at 20oC, calculate the solubility of the solute in moldm-3
[Pb = 207, N = 14, O = 16]
Bayanin Amsa
The problem provides the amount of lead (II) trioxonitrate (V) dissolved in 100g of water at a specific temperature. The question asks for the solubility of the solute in moldm-3. To calculate the solubility, we need to determine the amount of the solute (lead (II) trioxonitrate (V)) that is present in one liter of the solution. We can use the given amount of the solute and the molar mass of the compound to calculate the number of moles present in the solution. First, we calculate the molar mass of lead (II) trioxonitrate (V): 207 (mass of lead) + 3(16) (mass of oxygen) + 2(14) (mass of nitrogen) = 406 g/mol Next, we calculate the number of moles of the solute present in the solution: 16.55 g / 406 g/mol = 0.0407 mol Finally, we can calculate the solubility of the solute in moldm-3 by dividing the number of moles by the volume of the solution in liters: 0.0407 mol / 0.1 L = 0.407 moldm-3 Therefore, the solubility of the solute in moldm-3 is 0.407. The correct option is: - 0.05 g (none of the above) None of the options provided match the calculated solubility value.
Tambaya 24 Rahoto
Polystyrene is widely used as packaging materials for fragile objects during transportation because of its
Bayanin Amsa
Polystyrene is widely used as packaging materials for fragile objects during transportation because of its lightweight and low density. This means that it is easy to handle and does not add much weight to the package, making it more cost-effective to transport. Additionally, polystyrene has high compressibility, which means it can absorb impact and protect the fragile object inside the package from damage during transit.
Tambaya 25 Rahoto
The major and most effective way of controlling pollution is to
Tambaya 26 Rahoto
Ethyne is passed through a hot tube containing organo-nickel catalyst to produce
Bayanin Amsa
Passing ethyne (also known as acetylene) through a hot tube containing organo-nickel catalyst results in the production of benzene. This reaction is known as the "nickel carbide" or "cyclization" reaction. During the reaction, the ethyne molecules undergo a cyclization process in which the carbon atoms in the molecule bond together to form a ring. The organo-nickel catalyst helps to facilitate this process by providing a site for the carbon atoms to come together and form the ring. The reaction proceeds as follows: 2C2H2 → C6H6 In other words, two molecules of ethyne combine to form one molecule of benzene. Therefore, the correct answer is "benzene".
Tambaya 27 Rahoto
Beryllium and aluminium have similar properties because they
Bayanin Amsa
Beryllium and aluminium have some similar properties because they both belong to the same group of elements in the periodic table, known as Group 3A. Elements in the same group tend to have similar chemical and physical properties because they have the same number of electrons in their outermost shell, also known as valence electrons. This means they have similar reactivity and can form similar chemical bonds with other elements. However, beryllium and aluminium are not positioned diagonally to each other in the periodic table, and they also do not belong to the same period. Beryllium is in period 2, while aluminium is in period 3. They are both metals, but this is not the main reason for their similar properties.
Tambaya 29 Rahoto
The dispersion of a liquid in a liquid medium will give
Bayanin Amsa
When a liquid is dispersed in a liquid medium, it forms an emulsion. An emulsion is a type of mixture where small droplets of one liquid are suspended in another liquid. The two liquids are immiscible, which means they cannot dissolve in each other. In the case of an emulsion, one of the liquids is dispersed in the form of small droplets throughout the other liquid, forming a homogeneous mixture. This is different from a fog, which is a suspension of small liquid droplets in air, a gel, which is a solid or semi-solid material dispersed in a liquid, and an aerosol, which is a suspension of small solid or liquid particles in air.
Tambaya 30 Rahoto
The colour of litmus in a neutral medium is
Bayanin Amsa
Litmus is a type of natural dye that is extracted from certain lichens. It is used as an acid-base indicator in chemistry. When litmus is added to a neutral solution, it turns into a specific color. The color of litmus in a neutral medium is purple. Therefore, the correct answer to the question is "purple."
Tambaya 31 Rahoto
Water is said to be temporarily hard when it contains
Bayanin Amsa
Water is said to be temporarily hard when it contains Ca (HCO3)2 and Mg (HCO3)2 salts. Temporary hardness is caused by the presence of dissolved bicarbonate minerals in water, which can be removed by boiling. When the water is heated, the bicarbonate minerals decompose and release carbon dioxide gas, water, and solid calcium or magnesium carbonate. This solid material can be filtered out, leaving the water "soft."
Tambaya 32 Rahoto
If the difference in electronegativity of elements P and Q is 3.0. The bond that will be formed between them is
Bayanin Amsa
The difference in electronegativity between two elements determines the type of bond they will form. Electronegativity is the measure of an atom's ability to attract electrons towards itself in a covalent bond. If the difference in electronegativity between two elements is greater than 1.7, the bond between them is considered to be ionic. In an ionic bond, one atom will completely transfer its valence electrons to the other, resulting in the formation of ions with opposite charges that are held together by electrostatic forces. However, if the difference in electronegativity is less than 1.7, the bond between the elements is considered to be covalent. In a covalent bond, the atoms share valence electrons to fill their outermost energy levels. In the given question, the difference in electronegativity of elements P and Q is 3.0, which is greater than 1.7. Hence, the bond between them will be an ionic bond. Therefore, the answer is.
Tambaya 33 Rahoto
200 cm3 of a gas at 25oC exerts a pressure of 700 mmHg. Calculate its pressure if its volume increase 350 cm3 at 75oC.
Bayanin Amsa
\(\frac{p 1V 1}T 1\) = \(\frac{P 2V 2}{T 2}\) 700x200298 = \(\frac{P2X 350}{348}\) P2 = 700x200x348298x350 = 467.11mmHg
Tambaya 34 Rahoto
How many protons, neutrons and electrons respectively are present in the element 60/27 Co?
Bayanin Amsa
The element with the symbol 60/27 Co represents cobalt (Co) with an atomic mass of 60 and an atomic number of 27. The atomic number of an element represents the number of protons in the nucleus of an atom. Therefore, cobalt has 27 protons. The atomic mass of an element represents the number of protons and neutrons in the nucleus of an atom. To find the number of neutrons in cobalt, we can subtract the atomic number from the atomic mass: Number of neutrons = Atomic mass - Atomic number = 60 - 27 = 33 Therefore, cobalt has 33 neutrons. The number of electrons in an element is equal to the number of protons in the nucleus, assuming that the atom is neutral. Therefore, cobalt has 27 electrons. So, the correct answer is: 27, 33, and 27.
Tambaya 35 Rahoto
The alkanol obtained from the production of soap is
Bayanin Amsa
The alkanol obtained from the production of soap is glycerol. Glycerol is a byproduct of the saponification reaction, which is the process of making soap from fats or oils. The reaction between fats or oils and an alkaline solution produces soap and glycerol as the byproduct.
Tambaya 36 Rahoto
In a chemical reaction, the change in concentration of a reactant with time is
Bayanin Amsa
rate of reaction
Tambaya 37 Rahoto
The radio active radiation used in studying the arrangement of particles in giant organic molecules is
Tambaya 38 Rahoto
A gas X diffuses twice as fast as gas Y. if the relative molecular mass of X is 32, calculate the relative molecular mass of Y.
Bayanin Amsa
RYRX
= √My√My
1/2 = √32√y
32y
= 1/4
Y = 32x41
= 128
Tambaya 39 Rahoto
The basicity of CH3COOH is
Tambaya 40 Rahoto
In the electrolytic extraction of cacium from calcium chloride, the cathode is
Bayanin Amsa
Tambaya 41 Rahoto
Calculate the quantity of electricity in coulombs required to liberate 10g of copper from a copper compound[Cu=64, F = 96500 Cmol-1]
Bayanin Amsa
Cu → Cu2++ 2e
64g Cu requires 2 x 96500 coloumbs 10g Cu requires x columbs
x = 10x2x96500/64 =30156.3
Tambaya 42 Rahoto
The IUPAC nomenclature of the compound above is
Tambaya 43 Rahoto
In the extraction of sodium from fused sodium chloride, the anode is made of platinum because
Bayanin Amsa
The anode in the extraction of sodium from fused sodium chloride is made of platinum because sodium does not react with platinum. In this process, an electric current is passed through a molten mixture of sodium chloride to separate the sodium and chlorine ions. The anode attracts the negatively charged chloride ions, while the cathode attracts the positively charged sodium ions. The platinum anode does not react with the sodium ions, allowing for their separation from the chloride ions and the successful extraction of sodium.
Tambaya 44 Rahoto
The process of converting starch to ethanol is
Bayanin Amsa
The process of converting starch to ethanol is called "fermentation". Fermentation is a biological process that converts sugars, such as glucose, into ethanol and carbon dioxide. In the case of starch, it must first be hydrolyzed (broken down) into simpler sugars, like glucose, before it can be fermented. This hydrolysis is usually achieved through the addition of enzymes, such as amylases. Once the starch has been converted into simpler sugars, yeast or bacteria are added to initiate the fermentation process, in which the sugars are converted into ethanol and carbon dioxide. This process is widely used in the production of alcoholic beverages, such as beer and wine, as well as in the production of biofuels, such as ethanol.
Tambaya 45 Rahoto
The general formula of an alkyl halide (where X represents the halide) is?
Tambaya 46 Rahoto
Fluorine does not occur in the free state in nature because
Bayanin Amsa
Fluorine is a highly reactive element and readily reacts with other elements to form compounds. This high reactivity is why fluorine does not occur in the free state in nature. Instead, it is found in the earth's crust combined with other elements in compounds such as fluorite (calcium fluoride), cryolite (sodium aluminum fluoride), and fluorspar (calcium fluoride). The high reactivity of fluorine is due to its electron configuration, which makes it highly electronegative. This means that it has a strong tendency to attract electrons to itself, making it highly reactive and prone to forming chemical bonds with other elements. So, the reason fluorine does not occur in the free state in nature is due to its high reactivity, which corresponds to: of its high reactivity.
Tambaya 47 Rahoto
A few drops of NaOH solution was added to an unknown salt forming a white precipitate which is insoluble in excess solution. The cation likely present is
Bayanin Amsa
Tambaya 48 Rahoto
A compound that gives a brick-red colour to a non-luminous flame is likely to contain
Tambaya 49 Rahoto
What is the percentage of sulphur in sulphur (IV) oxide?
Bayanin Amsa
Sulfur (IV) oxide, also known as sulfur dioxide (SO2), is composed of one sulfur atom and two oxygen atoms. To determine the percentage of sulfur in sulfur (IV) oxide, we need to calculate the molar mass of the compound and then determine the molar mass of the sulfur atom. The molar mass of SO2 can be calculated by adding the atomic masses of sulfur and two oxygen atoms: Sulfur: 32.07 g/mol Oxygen: 2 x 16.00 g/mol = 32.00 g/mol Molar mass of SO2 = 32.07 g/mol + 32.00 g/mol = 64.07 g/mol The molar mass of sulfur is 32.07 g/mol, which means that sulfur makes up 32.07 g of the total molar mass of 64.07 g/mol. To determine the percentage of sulfur in SO2, we can use the following formula: percentage of sulfur = (mass of sulfur / total mass of compound) x 100% Substituting the values we obtained: percentage of sulfur = (32.07 g/mol / 64.07 g/mol) x 100% percentage of sulfur = 50% Therefore, the answer is option D, which is 50%.
